Problem: Solve for $x$, ignoring any extraneous solutions: $\dfrac{x^2}{x - 4} = \dfrac{2x + 8}{x - 4}$
Explanation: Multiply both sides by $x - 4$ $ \dfrac{x^2}{x - 4} (x - 4) = \dfrac{2x + 8}{x - 4} (x - 4)$ $ x^2 = 2x + 8$ Subtract $2x + 8$ from both sides: $ x^2 - (2x + 8) = 2x + 8 - (2x + 8)$ $ x^2 - 2x - 8 = 0$ Factor the expression: $ (x - 4)(x + 2) = 0$ Therefore $x = 4$ or $x = -2$ However, the original expression is undefined when $x = 4$. Therefore, the only solution is $x = -2$.